. Annals of Philosophy. 1819.] and Minima of Quantities. 195 Multiplying by x — a, and transposing, the equation becomes a8 — m x + a m = o ; .2 x — m = o and m = 2 x. But from the first equation m = .\ = 2 x, which * j — a * — a reduced gives x = 2 «. When radical quantities enter the proposed function, it does not appear that the common rule for reducing equations having two equal roots is generally applicable. We shall, therefore, investigate one by which examples of this kind maybe resolved. Since the equationy'jr = y has two affirmative roots, let/) = less = root and p + e = greater (e be
. Annals of Philosophy. 1819.] and Minima of Quantities. 195 Multiplying by x — a, and transposing, the equation becomes a8 — m x + a m = o ; .2 x — m = o and m = 2 x. But from the first equation m = .\ = 2 x, which * j — a * — a reduced gives x = 2 «. When radical quantities enter the proposed function, it does not appear that the common rule for reducing equations having two equal roots is generally applicable. We shall, therefore, investigate one by which examples of this kind maybe resolved. Since the equationy'jr = y has two affirmative roots, let/) = less = root and p + e = greater (e being their difference, and represented in the illustration of the lemma by the fine P P'). Then/) and p + e substituted for .r in the equation fx = y give the same result, viz. y. Therefore f p = y = f{p + e) (A). After developing the second member of" this equation, and taking away the quantities that are common to each side, all the remaining terms will be divisible by e, and we shall have an equation containing p, e, and constant quantities, which will be true for every value of the function. But when y = m, e = o ; therefore all the terms containing e and its powers will vanish, and we shall have an equation expressing the relation between p and given quantities, from the resolution of which, p or its equal X will be known. Let us apply this method to problem 2, where we have given x — x3 = y, to find # when y = m. This equation has two affirmative roots (by the lemma). Let jp = less root and/) + e = greater. Then these substituted for x in the proposed equation give the same result, viz. y. Therefore, p — p3 = y = p + e — (p + ef, or p — p3 = p + e — ps — 3 p? e — 3 p eQ — e3. Taking away p — p3 from each side, and dividing the remaining terms by e, we have 1 — 3 p* — 3 p e — ee = o. Now make e = o (because when y = m, e vanishes), and the last equation becomes 1 — 3 p- = o . = //i. = x. From this example, it is evident that in d
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